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Question

(B) Let the mirrors be $MN$ and $MP$ inclined at an angle $	heta$. $A$ light ray $AB$ is incident parallel to mirror $MN$ on mirror $MP$ at point $B$

Vedclass · Accepted Answer

$30^{\circ}$

Vedclass · Answer

(B) Let the mirrors be $MN$ and $MP$ inclined at an angle $	heta$. $A$ light ray $AB$ is incident parallel to mirror $MN$ on mirror $MP$ at point $B$.After the first reflection at $B$,the ray travels to point $C$ on mirror $MN$.After the second reflection at $C$,the ray travels to point $D$ on mirror $MP$.For the ray to retrace its path after the third reflection,it must strike the mirror $MP$ normally at point $D$.Thus,$CD \perp MP$,which implies $\angle CDM = 90^{\circ}$.In $	riangle MCD$,the sum of angles is $180^{\circ}$,so $\angle MCD = 180^{\circ} - 90^{\circ} - 	heta = 90^{\circ} - 	heta$.Since $CD$ is the reflected ray from $C$,the angle of incidence at $C$ is equal to the angle of reflection. Thus,$\angle BCM = \angle MCD = 90^{\circ} - 	heta$.In $	riangle BCM$,the sum of angles is $180^{\circ}$,so $\angle MBC = 180^{\circ} - \angle BCM - \angle BMC = 180^{\circ} - (90^{\circ} - 	heta) - 	heta = 90^{\circ}$.Since $AB \parallel MN$,the alternate interior angle $\angle ABC = \angle BCM = 90^{\circ} - 	heta$.Also,from the geometry at point $B$,$\angle ABC + \angle MBC + \angle PBA = 180^{\circ}$.Since $AB \parallel MN$,$\angle PBA = 	heta$ (corresponding angles).Substituting the values: $(90^{\circ} - 	heta) + 90^{\circ} + 	heta = 180^{\circ}$. This is always true for any $	heta$ if the ray hits $C$ at $90^{\circ} - 	heta$.However,for the specific condition of retracing after the third reflection,we require $\angle MBC = 90^{\circ}$.From $	riangle BCM$,$\angle MBC = 180^{\circ} - 	heta - (90^{\circ} - 	heta) = 90^{\circ}$.Using the property of reflection at $B$,the angle of incidence equals the angle of reflection. The angle between the incident ray $AB$ and the mirror $MP$ is $	heta$. The reflected ray $BC$ makes an angle $	heta$ with the mirror $MP$. Thus,$\angle PBC = 	heta$. Then $\angle ABC = 180^{\circ} - 2	heta$. Since $AB \parallel MN$,$\angle ABC + \angle BCM = 180^{\circ} \Rightarrow (180^{\circ} - 2	heta) + (90^{\circ} - 	heta) = 180^{\circ} \Rightarrow 3	heta = 90^{\circ} \Rightarrow 	heta = 30^{\circ}$.

Two mirrors are inclined at an angle $\theta$ as shown in the figure. $A$ light ray is incident parallel to one of the mirrors. The ray will start retracing its path after the third reflection if $\theta = ........$

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